Editing Simple continued fraction (section)

===Regular patterns in continued fractions===
While there is no discernible pattern in the simple continued fraction expansion of {{pi}}, there is one for {{math|''e''}}, the [[e (mathematical constant)|base of the natural logarithm]]:

:<math>e = e^1 = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, \dots],</math>

which is a special case of this general expression for positive integer {{mvar|n}}:

:<math>e^{1/n} = [1; n-1, 1, 1, 3n-1, 1, 1, 5n-1, 1, 1, 7n-1, 1, 1, \dots] \,\!.</math>

Another, more complex pattern appears in this continued fraction expansion for positive odd {{mvar|n}}:

:<math>e^{2/n} = \left[1; \frac{n-1}{2}, 6n, \frac{5n-1}{2}, 1, 1, \frac{7n-1}{2}, 18n, \frac{11n-1}{2}, 1, 1, \frac{13n-1}{2}, 30n, \frac{17n-1}{2}, 1, 1, \dots \right] \,\!,</math>

with a special case for {{math|1=''n'' = 1}}:

:<math>e^2 = [7; 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12, 54, 14, 1, 1 \dots, 3k, 12k+6, 3k+2, 1, 1 \dots] \,\!.</math>

Other continued fractions of this sort are

:<math>\tanh(1/n) = [0; n, 3n, 5n, 7n, 9n, 11n, 13n, 15n, 17n, 19n, \dots] </math>

where {{mvar|n}} is a positive integer; also, for integer {{mvar|n}}:

:<math>\tan(1/n) = [0; n-1, 1, 3n-2, 1, 5n-2, 1, 7n-2, 1, 9n-2, 1, \dots]\,\!,</math>

with a special case for {{math|1=''n'' = 1}}:

:<math>\tan(1) = [1; 1, 1, 3, 1, 5, 1, 7, 1, 9, 1, 11, 1, 13, 1, 15, 1, 17, 1, 19, 1, \dots]\,\!.</math>

If {{math|''I''<sub>''n''</sub>(''x'')}} is the modified, or hyperbolic, [[Bessel function]] of the first kind, we may define a function on the rationals {{sfrac|''p''|''q''}} by

:<math>S(p/q) = \frac{I_{p/q}(2/q)}{I_{1+p/q}(2/q)},</math>

which is defined for all rational numbers, with {{mvar|p}} and {{mvar|q}} in lowest terms. Then for all nonnegative rationals, we have

:<math>S(p/q) = [p+q; p+2q, p+3q, p+4q, \dots],</math>

with similar formulas for negative rationals; in particular we have

:<math>S(0) = S(0/1) = [1; 2, 3, 4, 5, 6, 7, \dots].</math>

Many of the formulas can be proved using [[Gauss's continued fraction]].